Water is being boiled in an open kettlethat has a (L = 0.500-cm) -thick circular aluminum bottle with a radius of (r = 12.0cm).

If the water boils away at a rate of

\(\displaystyle{\frac{{{d}{m}}}{{{\left.{d}{t}\right.}}}}={0.500}\ {k}\frac{{g}}{\min}={8.33}\cdot{10}^{{-{3}}}\ {k}\frac{{g}}{{s}}\)

Let the temperature of the lower surface of the bottom of thekettle be \(\displaystyle{{t}_{{2}}^{\circ}}\) C

The temperature of thetop surface of the bottom of the kettle (\(\displaystyle{t}_{{1}}={100}^{\circ}{C}\))

Since the heat given by the aluminium bottle = heattaken by boiled water

\(\displaystyle{k}{A}{\left({t}_{{2}}-{t}_{{1}}\right)}\cdot{\frac{{{\left.{d}{t}\right.}}}{{{L}}}}{)}={d}{m}\cdot{L}_{{s}}\)

Where

\(\displaystyle{K}={238}\ \frac{{J}}{{s}}\)

\(\displaystyle{A}={p}{r}^{{2}}\)

dt=time taken for the mass dm

Latent Heat of steam (\(\displaystyle{L}_{{s}}\))=2256000 J/kg

Ahain from above equation

\(\displaystyle{k}{A}{\frac{{{t}_{{2}}-{t}_{{1}}}}{{{d}}}}={\frac{{{d}{m}}}{{{\left.{d}{t}\right.}}}}\cdot{L}_{{s}}\)

Now calculate the value of \(\displaystyle{t}_{{2}}\) from above

If the water boils away at a rate of

\(\displaystyle{\frac{{{d}{m}}}{{{\left.{d}{t}\right.}}}}={0.500}\ {k}\frac{{g}}{\min}={8.33}\cdot{10}^{{-{3}}}\ {k}\frac{{g}}{{s}}\)

Let the temperature of the lower surface of the bottom of thekettle be \(\displaystyle{{t}_{{2}}^{\circ}}\) C

The temperature of thetop surface of the bottom of the kettle (\(\displaystyle{t}_{{1}}={100}^{\circ}{C}\))

Since the heat given by the aluminium bottle = heattaken by boiled water

\(\displaystyle{k}{A}{\left({t}_{{2}}-{t}_{{1}}\right)}\cdot{\frac{{{\left.{d}{t}\right.}}}{{{L}}}}{)}={d}{m}\cdot{L}_{{s}}\)

Where

\(\displaystyle{K}={238}\ \frac{{J}}{{s}}\)

\(\displaystyle{A}={p}{r}^{{2}}\)

dt=time taken for the mass dm

Latent Heat of steam (\(\displaystyle{L}_{{s}}\))=2256000 J/kg

Ahain from above equation

\(\displaystyle{k}{A}{\frac{{{t}_{{2}}-{t}_{{1}}}}{{{d}}}}={\frac{{{d}{m}}}{{{\left.{d}{t}\right.}}}}\cdot{L}_{{s}}\)

Now calculate the value of \(\displaystyle{t}_{{2}}\) from above